$\begingroup$You are overcounting. For example, $112344$ gets counted four times, once for each position of a $1$ being among the first four digits times once for each position of a $4$ being among the first four digits. It is probably easier to count all the six digit numbers with these digits and subtract off the ones that don't have all the digits. You will need the inclusion-exclusion principle to do that. There are many examples on the site.$\endgroup$
– Ross MillikanNov 22 '17 at 20:41

$\begingroup$@ShakedBader I think it will be more complicated. See 123432 it wil be counted at least two times.$\endgroup$
– hemuNov 22 '17 at 20:43

1

$\begingroup$Inclusion Exclusion is a good way to proceed. Compute the number with no restrictions, then subtract those missing a specified digit, then add back those missing a specified pair, and so on.$\endgroup$
– luluNov 22 '17 at 20:52